# Why the electric potential inside a conductor doesn't equal zero?

Assume that a negative charge in the space near of it a small positive charge, Then for sure that the negative charge will attract the positive charge naturally because it has an electric field around it ,So the point which the positive charge at will have an electric potential that equal to $$V = \frac {kq}{r}$$.

But, If there is no any electric field in a point then the force must be zero because $$E = \frac {F}{q}$$, Since $$E = 0$$ then the force must be zero, So there is no any potential energy provided to any point around.

When I begin to study electric potential inside a charged hollow or solid sphere I really surprised, How is this? There is electric potential without electric field ! Can any one explain this for me?

• You can have any potential, that's meaningless. Your formula is wrong. For a single charge, $V=Kq/r + constant$, and you can have any constant. A constant potential does nothing. It's differences of potential what create electric fields. Apr 13 '19 at 17:26
• @FGSUZ , Could you please explain more ? I cant understand . Apr 13 '19 at 17:32
• Just to add to @FGSUZ comment: if the potential were zero inside the conductor and non-zero outside there would be an infinite electric field at the boundary, which is unphysical. Because of that when solving for the potential we impose continuity to it. Apr 13 '19 at 18:09
• @ErickShock ,What caused the electric potential inside the conductor ? Apr 13 '19 at 18:15
• @Mohammad Alshareef because there are charges outside that caused a rearrangement of the internal charges of the conductor (or you added extra charges to the neutral conductor). The potential is influenced by all charges. Apr 13 '19 at 18:29

Okay, I'll ellaborate my comment with the purpose of making it clearer. It's true it was very chaotic, but that's because I wanted to address many thing in few words.

I think there are many issues here. I'll try to clear them up.

1.- Potential energies can start wherever you want.

The other answers have already pointed this out with a nice example: a ball on a platform. This is because we are more used to gravitational potential energy: $$E_{pg}=mgh$$.

We are assuming that we are on usual situations on Earth, and we consider that $$g$$ doesn't vary.

So, imagine we have a ball on the ground. We would say that its potential energy is $$0$$ because it is at $$0m$$ height.

However, now someone excavates a big hole $$10m$$ down in the ground. What if the ball falls into the hole?

You have two options:

a) The new potential energy will be negative: $$m\cdot g\cdot (-10m)$$. We maintain that the ball was originally at $$0m$$.

b) We redefine our origin of coordinates. Now the hole's bottom is height=0. Consequently, the ball was at $$h=+10m$$ before falling.

Well, the thing is that both options are absolutely valid. You can choose any of those options, and the energy will keep being conserved. All physical laws and processes will behave exactly the same way.

As a consequence: we can choose where we start counting the potential energy. We can decide where we have $$E_p=0$$. This does not happen to kinetic energy, only potential energy.

2.- The origin of optential energies MUST in fact be DECIDED BY YOU

There is no "special point" where the potential energy (PE) has to be 0. Nothing can force you to choose any special point. It's you who must decide where to set $$E_p=0$$.

This means you have freedom to choose that. This can be good or bad, but that's how it is.

3.- Potentials are relative quantities, not absolute

The electric potential energy of a point charge is not

$$V=K \frac{q}{r}$$

That would be quite absolute. You cannot actually get an absolute potential. What you can obtain is potential differences. The real formula you can obtain is:

$$V=\left( K \frac{q}{r} - K \frac{q}{r_0} \right) =Kq \left(\frac{1}{r}-\frac{1}{r_0}\right)$$

Where $$r_0$$ is the point you chose as reference. This means that you cannot calculate any "absolute potential", rather, you calculate "the difference of potentials between your point and the one you chose as reference". You can only caculate differences between 2 points.

And what's that $$r_0$$? The one you chose! Just like you choose what height is $$E_{pg}=0$$.

4.- Usual gauge

What we normally do is: we set $$r_0$$ to be at infinity. That means stablishing that $$V=0$$ at infinity. It also implies that $$V\neq0$$ unless the particle is infinitely far away from us.

If we chose this one, then the secnod term vanishes, and so

$$V==Kq \left(\frac{1}{r}-\frac{1}{r_0}\right) \rightarrow K\frac{q}{r}$$

This is the usal formula. But you must keep in muind that "points do not have an absolute potential". The potential depends on where you chose to put $$V=0$$. Only if you put $$V=0$$ at infinity you have this formula.

You cannot use this formula if therea re charges at infinity.

By the way, check that, if you choose another $$r_0$$, you will have

$$V=Kq \left(\frac{1}{r}-const\right) = Kq/r + constant$$

5.- Differences of potential are what matters.

The electric field is

$$E= dV/dx$$

and $$V=Kq/r + constant$$.

when you derive, the constant goes out. The electric field does not depend on where you chose the origin, as it must be.

In the same way, if you let a ball fall 10m high, you will have $$\Delta E_p= mg\Delta h$$, and that $$\Delta E_p$$ will become KE. Only differences of energy matter. It doesn't matter if the ball falls from 90m to 80m, or from 0m to minus 10m. Only differences of potential matter.

Same thing here. The force cannot depend on where you chose the origin, and it does not depend. Derivatives cancel the constant.

In conclusion:

There can be a potential without a field. A field is 0 if the potential is constant (derivative of a constant=0). The value of that constant depends on you. However, if tehre were a potential differnece, then the derivative would not be 0, and so you would have a field.

I'll end with a wise sentence from a wise teacher: Energy is free, what you pay for is DIFFERENCS OF ENERGY. Do you want to have 1000 joules right now? Okay, now you have. I just moved the origin. That's free. If you wanted to CHANGE the value with a fixed origin, then that's adding a potential difference, and that's work done, that's expensive. Work costs money, energy does not, as it depends on the origin. If energy weren't free, then you can be sure that the electric company would have a term on the bill called "PE origin", and of course, it would be more and more expensive every month.

As per @FGSUZ's answer, it's not the electric potential that matters: it's the potential difference.

Think of it in these terms: Let's say you have a ball on a horizontal platform, 1 meter above the ground. The ball represents a charge inside the conductor. The horizontal platform represents the conductor. The ball certainly has gravitational potential energy - it's above the ground. However, as long as the ball is only on the platform, it can't change its total energy (assuming you can't give the ball a push horizontally). The total energy will stay E = mgh, as that potential energy can't be converted into kinetic energy as long as the ball is on the platform.

Same thing here. As long as the charge is in the conductor, it's basically on that platform. There may be an electric potential (a positive height, in terms of the earlier analogy), but there is no electric potential difference (the platform is horizontal, not slanted; there is no change in elevation).

• @lan Ng , So do I need to work to bring a charge inside the conductor ? Apr 13 '19 at 18:17
The electric field is the negative gradient of the electric potential so one can have no gradient (i.e. no slope) but still a non-zero potential at some constant value $$V_0$$ in some region.
It is also perfectly possible to have a plateau at some height. On the plateau the slope is $$0$$ but the plateau need not be a sea-level. In this analogy the potential would be high and non-zero but the field would still be $$0$$.
In other words, it is perfectly possible for a region of space to be at the same non-zero potential; since the field is related to the change in potential, the field in that region would be $$0$$ even though the potential is not.